Question

Solve $\int \frac{\sin 4x}{\sin x}dx$.

Answer 1

$I=\int \frac{sin4x}{sinx}dx=\int \frac{2sin2xcos2x}{sinx}dx$

$=2\int \frac{2sinxcosx(2co{s}^{2}x-1)}{sinx}dx$

$\therefore sin2x=2sinxcosx$

$cos2x=2co{s}^{2}x-1$

$I=4\int cosx[1-2si{n}^{2}x]dx$

$sinx=t$

$cosxdx=dt$

$I=4\int (1-2t^2)dt$

$I=4[t-2\frac{t^3}{3}]+c$

$I=4[sinx-\frac{2}{3}si{n}^{3}x]+c$